Why does int pointer '++' increment by 4 rather than 1?
As you said, an int pointer points to an int. An int usually takes up 4 bytes and therefore, when you increment the pointer, it points to the "next" int in the memory - i.e., increased by 4 bytes. It acts this way for any size of type. If you have a pointer to type A, then incrementing a A* it will increment by sizeof(A).
Think about it - if you only increment the pointer by 1 byte, than it will point to a middle of an int and I can't think of an opportunity where this is desired.
This behavior is very comfortable when iterating over an array, for example.
Pointers are increased by the size of the type they point to, if the pointer points to char, pointer++ will increment pointer by 1, if it points to a 1234 bytes struct, pointer++ will increment the pointer by 1234.
This may be confusing first time you meet it, but actually it make a lot of sense, this is not a special processor feature, but the compiler calculates it during compilation, so when you write pointer+1 the compiler compiles it as pointer + sizeof(*pointer)
When you increment a T*, it moves sizeof(T) bytes.† This is because it doesn't make sense to move any other value: if I'm pointing at an int that's 4 bytes in size, for example, what would incrementing less than 4 leave me with? A partial int mixed with some other data: nonsensical.
Consider this in memory:
[↓ ]
[...|0 1 2 3|0 1 2 3|...]
[...|int |int |...]
Which makes more sense when I increment that pointer? This:
[↓ ]
[...|0 1 2 3|0 1 2 3|...]
[...|int |int |...]
Or this:
[↓ ]
[...|0 1 2 3|0 1 2 3|...]
[...|int |int |...]
The last doesn't actually point an any sort of int. (Technically, then, using that pointer is UB.)
If you really want to move one byte, increment a char*: the size of of char is always one:
int i = 0;
int* p = &i;
char* c = (char*)p;
char x = c[1]; // one byte into an int
†A corollary of this is that you cannot increment void*, because void is an incomplete type.